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Limits using Sandwich Theorem and L’Hospital’s Rule

Table of Content

Sandwich Theorem
We now discuss certain illustrations involving greatest integer functions
L’Hospital’s Rule
Related Resources

Sandwich Theorem

Sandwich theorem is an important concept of limits. It is often termed as the Squeeze theorem, Pinching Theorem or the Squeeze Lemma. The Squeeze principle is generally used on limit problems where the usual algebraic methods like the algebraic or the factorization methods fail in computing limits. Hence, in such a case the sandwich or the squeeze theorem tries to squeeze our problem in between the limits of two simple functions whose limits can be evaluated with ease and are in fact equal. In fact, this is the reason behind the name of this theorem.

Let f(x), g(x) and h(x) be three real numbers having a common domain D such that f(x) ≤ h(x) ≤ g(x) ∀ x in some interval containing the point c. Then,

If lim _x→a f(x) = lim _x→a g(x) = l, then lim _x→a h(x) = l.

This is known as Sandwich Theorem.

Illustration:

Examine lim _x→0 √(1 – cos x)/x

Solution:

solution6

View the following video for more on Sandwich Theorem

We now discuss certain illustrations involving greatest integer functions

Illustration 1:

lim _x→3 ([x] – 3)/(x – 3)

Solution:

lim _x→3 ([x] – 3)/(x – 3)

Towards the right of x = 3, [x] = 3

⇒ [x] – 3 = 0, in the right neighborhood of x = 3

⇒ lim _x→3+0 ([x] – 3)/(x – 3) = 0

Towards the left of x = 3, [x] = 2

⇒ [x] – 3 = -1, in the left neighborhood of x =3

⇒ lim _x→3+0 ([x] - 3)/(x – 3) = lim _x→3+0 (-1)/(x – 3) = ∞.

Thus lim _x→3 ([x] – 3)/(x – 3) does not exist.

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Illustration 2:

Prove that lim _n→∞ x⁵ [1/x³] = 0, where [.] denotes the greatest integer function.

Solution:

We know that (x - 1) < |x| ≤ x ∀ x ∈ R

So, 1/x³ -1 < [1/x³] ≤ 1/x³

⇒ x⁵(1/x³ -1) < x⁵ [1/x³ ] ≤ x⁵ (1/x³ ) for x > 0

And lim _x→0 x² - x⁵ = lim _x→0 x² = 0

⇒ lim _x→0⁺ x⁵ [1/x³ ] = 0

Also, x⁵ (1/x³) ≤ x⁵ [1/x³ ] < x⁵ (1/x³ -1) for x < 0

⇒ lim _x→0^- x⁵ [1/x³] = 0 => lim_x→0 x⁵ [1/x³ ] = 0.

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Illustration 3:

Evaluate lim _n→∞ ([x] + [2x] + [3x] + ...........= [nx]) / n², where [.] denotes the greatest integer function.

Solution:

We know that (x - 1) < x

⇒ x + 2x + .......... + nx – n < ∑_r=1ⁿ [rx] ≤ x + 2x + ......... + nx

⇒ (x.{n (n + 1)/2} - n) < ∑_r=1ⁿ [rx] ≤ {x.n (n + 1)/2}

⇒ x/2 (1 + 1/n) – 1/n < 1/n² ∑_r=1ⁿ [rx] ≤ x/2 (1 + 1/n)

Now, lim _n→∞ x/2 (1 + 1/n) = x/2 and lim _n→∞ x/2 (1 + 1/n) – 1/n = x/2.

Using Sandwich theorem we find that lim _n→∞ ([x] + [2x] + ..........+ [nx])/n² = x/2

L’Hospital’s Rule

The L Hospital’s Rule is used in cases where the limit obtained is in the form of 0/0, ∞/∞, ∞ – ∞, 0, ∞, 0⁰, ∞⁰, 1^∞. The other indeterminate forms are ∞-∞,0,∞,00,∞0,1∞. These are known as indeterminate forms. According to this rule,

Let f(x) and g(x) be functions differentiable in the neighborhood of the point a, except may be at the point a itself. If lim _x→a f(x) = 0 = lim_x→ag(x) or lim _x→af(x) = ∞ = lim _x→a g(x), then

lim _x→a f(x)/g(x) = lim _x→a f ’(x)/g'(x) provided that the limit on the right either exists as a finite number or is ± ∞ .

The following illustrations describe the rule:

Illustration 1:

Evaluate lim_x→1 (1 – x + ln x)/(1 + cos πx)

Solution:

lim _x→1 (1 – x + ln x)/(1 + cos πx) (of the form 0/0)

= lim _x→1 (1 – 1/x)/(-π sin πx) (still of the form 0/0)

= lim _x→1 (x – 1)/(πx sin πx) (algebraic simplification)

= lim _x→1 1/(πx sin πx + π²x cos πx ) (L' Hospital's rule again)

= - 1/π²

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Illustration 2:

Let f : R→R be such that f(1) = 3 and f’(1) = 6. Then lim _x→0 f(1+x)/f(1))^1/x equals (IIT JEE 2002)

(a) 1 (b) e^1/2

Solution:

The given function is y = [f(1 + x)/ f(1)]^1/x

Taking log both sides we have,

log y = 1/x [log f(1 + x) – log f(1)]

Taking limit on both sides we get,

lim _x→0log y = lim _x→0[f’(1 + x) / f(1 + x)] (using Hospital’s rule)

= f’(1)/ f(1)

= 6/3

log (lim _x→0 y) = 2.

lim _x→0y = e² .

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Illustration 3:

Evaluate lim x→y (xy - yx)/(xx - yy).

Solution:

lim x→y (xy - yx)/(xx - yy) which is a [0/0] form

= lim _x→y (yxy-1 - yx log y)/(xx log (ex))

= (1 – log y)/ log (ey)

Q1. In Sandwich theorem, the function f(x) ≤ h(x) ≤ g(x) ∀ x in some interval containing the point c. Then,

(a) c may be a finite number, +∞ or – ∞.

(b) c can only be a finite number

(d) none of these.

Q2. L’Hospital’s Rule can be used in solving limits

(a) c may be a finite number, +∞ or – ∞.

(b) of indeterminate forms.

(d) none of these.

Q3. The limit L as obtained in Sandwich theorem

(a) L has to be positive

(b) L can only be a finite number

(d) L may be a finite number, +∞ or – ∞.

Q4. Using L’Hospital’s Rule, we have lim_x→1 [f(x)/g(x)] =

(a) f(x)/g’(x)

(b) f’(x)/g(x)

(d) none of these.

Q1.	Q2.	Q3.	Q4.
(a)	(b)	(d)	(c)

Related Resources

You may wish to refer Differentiability.
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For getting an idea of the type of questions asked, refer the previous year papers.

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